SOLUTION: I have some factoring problems.. I have solved some, but the others just dont work. 3x(a-3)-5(3-a) I don't even know where to start, it looks factored to me? 20x^2-7x-6. I h

Question 342932: I have some factoring problems.. I have solved some, but the others just dont work.
3x(a-3)-5(3-a) I don't even know where to start, it looks factored to me?
20x^2-7x-6. I have tried about every different combination I can for foil for this one, it just does not work itself out.. It is driving me CRAZY!!!
4x^4 + 12x^3- 28x^2. I don't even know where to start to be honest. Is this a foil problem?
Thank you for your time..I can not tell you how valuable this site has been to me.
Found 3 solutions by stanbon, nyc_function, akc:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
3x(a-3)-5(3-a)
---
3x(a-3)+5(a-3)
---
(a-3)(3x+5)
----------------------------------
20x^2-7x-6.
---
20x^2-15x+8x-6
---
5x(4x-3)+2(4x-3)
---
(4x-3)(5x+2)
========================
4x^4 + 12x^3- 28x^2
---------
4x^2[x^2+3x-7]
==================
Cheers,
Stan H.
============

Answer by nyc_function(2741)   (Show Source): You can put this solution on YOUR website!
The expression 3x(a-3)-5(3-a) becomes 3x(a-3)-5(3-a) after using the distributive rule.

20x^2 - 7x - 6 factors out to be (5x+2)(4x-3). To factor this one, I used the GROUPING METHOD.

4x^4 + 12x^3- 28x^2 factors out to be 4x^2(x^2 + 3x - 7). To factor this one, I needed to do some extra substitutions in terms of letting y = x^2.

Done!

Answer by akc(7)   (Show Source): You can put this solution on YOUR website!
Please post one question is a post.
1.
3x(a-3)-5(3-a) (factorisaton means there should not be + or - between two terms)
=3x(a-3)-5(-(a-3)) [-(a-3)= 3-a]
=3x (a-3)+ 5 (a-3) [two - equals +]
=(a-3)(3x + 5) [taking (a-3) common from both terms]
(a-3) & (3x+5) are the factors.
.
.
.
.
2.

=
=
=
= (4x-3)(5x + 2)
(4x-3)&(5x + 2) are the factors
.
.
.
.
3.
=
& are the factors.

Again plz ask single question at a time. you can email me for immediate help. my email is akc1@uno.edu
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